Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iostream>
#include<vector>
#include<queue>
using namespace std;

void dijkstra(int from, vector<int>* v, vector<int> &d){
    queue<int> q;
    q.push(from);
    d[from] = 0;
    while(!q.empty()){
        int x = q.front();  q.pop();
        for(int next : v[x]){
            if(d[next] != 1<<30)    continue;
            d[next] = d[x]+1;
            q.push(next);
        }
    }
}

typedef long long ll;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    s--, t--;
    
    vector<int> v[n];
    while(m-- > 0){
        int x, y;
        cin >> x >> y;
        x--, y--;
        v[x].push_back(y);
        v[y].push_back(x);
    }

    vector<int> a(n, 1<<30), b(n, 1<<30);
    dijkstra(s, v, a);
    dijkstra(t, v, b);

    vector<int> cnta(n+1, 0), cntb(n+1, 0);
    for(int i = 0; i < n; i++)  if(a[i] != 1<<30)   cnta[a[i]]++;
    for(int i = 0; i < n; i++)  if(b[i] != 1<<30)   cntb[b[i]]++;

    ll ans = 0;
    for(int i = 0; i <= a[t]-2; i++){
        ans += (ll)(cnta[i]) * cntb[a[t]-2-i];
    }
    cout << ans << endl;

    return 0;
}